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Math Help - evaluating solutions to trig equation??

  1. #1
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    evaluating solutions to trig equation??

    <br />
\begin{array}{l}<br />
 {\rm Trying to find all solutions to interval inclusive of  - pi to  + pi for } \\ <br />
 2 + \cos (2x) = 3\cos (x) \\ <br />
 {\rm What I have done is this:} \\ <br />
 2 + [\cos ^2 (x) - \sin ^2 (x)] = 3\cos (x) \\ <br />
 2 + \cos ^2 (x) - \sin ^2 (x) = 3\cos (x) \\ <br />
 2 + \cos ^2 (x) - 3\cos (x) = \sin ^2 (x) \\ <br />
 \end{array}

    <br />
\begin{array}{l}<br />
 2 + \cos ^2 (x) - 3\cos (x) = 1 - \cos ^2 (x) \\ <br />
 2 + 2\cos ^2 (x) - 3\cos (x) = 1 \\ <br />
 2\cos ^2 (x) - 3\cos (x) + 1 = 0 \\ <br />
 (2\cos (x) - 2)(\cos (x) - 1) = 0 \\ <br />
 {\rm So  then  }2\cos (x) = 2{\rm   and  }\cos (x) = 1 \\ <br />
 {\rm Which gives one root of cos(x) = 1} \\ <br />
 \end{array}
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Craka View Post
    <br />
\begin{array}{l}<br />
 {\rm Trying to find all solutions to interval inclusive of  - pi to  + pi for } \\ <br />
 2 + \cos (2x) = 3\cos (x) \\ <br />
 {\rm What I have done is this:} \\ <br />
 2 + [\cos ^2 (x) - \sin ^2 (x)] = 3\cos (x) \\ <br />
 2 + \cos ^2 (x) - \sin ^2 (x) = 3\cos (x) \\ <br />
 2 + \cos ^2 (x) - 3\cos (x) = \sin ^2 (x) \\ <br />
 \end{array}

    <br />
\begin{array}{l}<br />
 2 + \cos ^2 (x) - 3\cos (x) = 1 - \cos ^2 (x) \\ <br />
 2 + 2\cos ^2 (x) - 3\cos (x) = 1 \\ <br />
 2\cos ^2 (x) - 3\cos (x) + 1 = 0 \\ <br />
 (2\cos (x) - 2)(\cos (x) - 1) = 0 \\ <br />
 {\rm So  then  }2\cos (x) = 2{\rm   and  }\cos (x) = 1 \\ <br />
 {\rm Which gives one root of cos(x) = 1} \\ <br />
 \end{array}
    Something went wrong in your factorisation

    (2 \cos(x)-2)(\cos(x)-1)=2(\cos(x)-1)^2=2 \cos^2(x)-{\color{red}4}\cos(x)+{\color{red}2}

    2 \cos^2(x)-3\cos(x)+1=(2 \cos(x)-{\color{magenta}1})(\cos(x)-1)=0

    This gives \cos(x)=\frac 12 or \cos(x)=1.
    Last edited by Moo; August 22nd 2008 at 08:45 AM.
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  3. #3
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    Hello, Craka!

    There are three forms for \cos2\theta

    . . \cos2\theta \;=\;\begin{Bmatrix}\cos^2\!\theta - \sin^2\!\theta \\ 2\cos^2\!\theta - 1 \\ 1 - 2\sin^2\!\theta \end{Bmatrix}


    The last two can be derived from the first,
    but I've found it expedient to memorize all three.


    Find all solutions on [-\pi,\:+\pi]\!:\;\;2 + \cos (2x) \:=\:3\cos (x)

    We have: . 2 + \cos2x \:=\:3\cos x

    . . . . . 2 + \overbrace{2\cos^2\!x - 1} \:=\:3\cos x

    . . 2\cos^2\!x - 3\cos x + 1 \:=\:0

    . (\cos x - 1)(2\cos x - 1) \:=\:0


    Therefore: . \begin{array}{ccccc}\cos x - 1 \:=\:0 & \Rightarrow & \cos x \:=\:1 & \Rightarrow &\boxed{x \:=\:0} \\ \\[-3mm]<br />
2\cos x - 1 \:=\:0 & \Rightarrow & \cos x \:=\:\frac{1}{2} & \Rightarrow & \boxed{x \:=\:\pm\frac{\pi}{3}} \end{array}

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